Re: Re: Simplify with NestedLessLess?

*To*: mathgroup at smc.vnet.net*Subject*: [mg106581] Re: [mg106531] Re: [mg106487] Simplify with NestedLessLess?*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sun, 17 Jan 2010 07:10:53 -0500 (EST)*References*: <201001141049.FAA19892@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

You said C<<Cf, so let's say C = delta CF, where delta is small. We compute a series around delta=0 and, after giving it a look, we set delta = 0. expr = Normal@Series[-Cf^2 L2^2 Rg^2 Vg^4 + 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg) /. C -> delta Cf, {delta, 0, 6}] firstTry = expr /. delta -> 0 12 Cf Rg^2 Vd^2 - Cf^2 L2^2 Rg^2 Vg^4 + delta (12 Cf Rg^2 Vd^2 + 6 Cf Rg^2 Vd Vg) 12 Cf Rg^2 Vd^2 - Cf^2 L2^2 Rg^2 Vg^4 In a more complicated situation (if a few Series terms were unconvincing), we might use Limit, instead: expr = -Cf^2 L2^2 Rg^2 Vg^4 + 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg) /. C -> delta Cf secondTry = Limit[expr, delta -> 0] -Cf^2 L2^2 Rg^2 Vg^4 + 3 (4 Cf Rg^2 Vd^2 + 4 Cf delta Rg^2 Vd^2 + 2 Cf delta Rg^2 Vd Vg) Cf Rg^2 (12 Vd^2 - Cf L2^2 Vg^4) In this case, both answers are the same: firstTry == secondTry // Simplify True Bobby On Sat, 16 Jan 2010 13:38:22 -0600, Dave Bird <dbird at ieee.org> wrote: > Interesting! But, I don't think I am correctly communicating what I'm > after yet. (Although, I admit that I am struggling some to keep up with > you guys in your Mathematica replies due to my inexperience.) > > The original expression that I put up for illustration is: > > -Cf^2 L2^2 Rg^2 Vg^4+3 (4 C Rg^2 Vd^2+4 Cf Rg^2 Vd^2+2 C Rg^2 Vd Vg) > > We compare 4 C Rg^2 Vd^2 to 4 Cf Rg^2 Vd^2 because the two terms share > common coefficients so that they "reduce" to (4 Rg^2 Vd^2+4 Rg^2 Vd^2) > (C+Cf) . Thus it becomes obvious that C may be discarded w.r.t. Cf. > > Please forgive if I have missed the correct application of your > suggestion, and thanks for the interest. > > Dave > > DrMajorBob wrote: >> Series[-Cf^2 L2^2 Rg^2 Vg^4 + >> 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg), {C, 0, >> 5}] // Simplify >> >> SeriesData[C, 0, { >> Cf Rg^2 (12 Vd^2 - Cf L2^2 Vg^4), 6 Rg^2 Vd (2 Vd + Vg)}, 0, 6, 1] >> >> Bobby >> >> On Fri, 15 Jan 2010 02:21:09 -0600, Dave Bird <dbird at ieee.org> wrote: >> >>> Not infinitesimals. I'm working in analog circuit design/analysis. I >>> have a 3 pole symbolic circuit response (third order) which is not >>> easily separable. I can use Mathematica to find the three roots of the >>> response. But, the roots are, of course, very messy. I know that >>> certain >>> elements in the circuit are orders of magnitude larger than other like >>> elements - capacitors in this case. For example, one small section of >>> one root is >>> >>> -Cf^2 L2^2 Rg^2 Vg^4+3 (4 C Rg^2 Vd^2+4 Cf Rg^2 Vd^2+2 C Rg^2 Vd Vg) >>> >>> I know that C<<Cf. By careful inspection, I can see that the first term >>> in the parens will drop out compared to the second term in the parens. >>> I >>> would like Mathematica to do this without my having to examine it so >>> closely since there are many other like situations. >>> >>> This kind of situation occurs in many other engineering situations. >>> >>> Hope this helps clarify. >>> >>> Thanks for the interest. >>> >>> Dave >>> >>> >>> >>> >>> Daniel Lichtblau wrote: >>>> Dave Bird wrote: >>>>> Thanks Daniel for the observation. I forgot to add that both a, and b >>>>> are real positive. That, of course would have to be added to the >>>>> assumptions. >>>>> >>>>> Dave >>>> >>>> It's still not obvious what you are wanting to do. I have the idea you >>>> are working in some sense with infinitesmals. If so, I doubt Simplify >>>> would be the best tool for removing them; it really can only do that >>>> if it is told, in some way, to replace them with zero. How might one >>>> instruct Simplify to figure that out? >>>> >>>> Daniel >>>> >>>> >>>>> Daniel Lichtblau wrote: >>>>>> dbird wrote: >>>>>>> Please excuse if this has been answered before, but I can't find >>>>>>> it. >>>>>>> >>>>>>> Is there some way to do a Simplify with assumptions using a >>>>>>> NestedLessLess or something similar? For example: >>>>>>> >>>>>>> d=a+b >>>>>>> Simplify[d,NestedLessLess[a,b]] >>>>>>> >>>>>>> Answer is: >>>>>>> a+b >>>>>>> >>>>>>> Answer should be: >>>>>>> b >>>>>>> >>>>>>> Thanks, >>>>>>> >>>>>>> Dave >>>>>> >>>>>> I fail to see why the result should be b. >>>>>> >>>>>> Daniel Lichtblau >>>>>> Wolfram Research >>>>>> >>>>>> >>>> >>>> >>> >> >> -- DrMajorBob at yahoo.com

**References**:**Simplify with NestedLessLess?***From:*dbird <dbird@ieee.org>

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**Re: Re: Simplify with NestedLessLess?**

**Re: Re: Simplify with NestedLessLess?**

**Re: Re: Simplify with NestedLessLess?**